Answer:
12.36 L.
Explanation:
We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:
Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]
= 207.2 + 2[14.01 + 48]
= 207.2 + 2[62.01]
= 207.2 + 124.02
= 331.22 g/mol
Mass of Pb(NO₃)₂ = 147.1 g
Mole of Pb(NO₃)₂ =?
Mole = mass / Molar mass
Mole of Pb(NO₃)₂ = 147.1 / 331.22
Mole of Pb(NO₃)₂ = 1.104 moles.
Next, we shall determine the number of mole of oxygen gas, O₂, produce from the reaction. This can be obtained as follow:
2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂
From the balanced equation above,
2 moles of Pb(NO₃)₂ decomposed to produce 1 mole of O₂.
Therefore, 1.104 moles of Pb(NO₃)₂ will decompose to produce = (1.104 × 1)/2 = 0.552 mole of O₂.
Finally, we shall determine the volume occupied by 0.552 mole of oxygen gas, O₂. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.552 mole of O₂ will occupy = 0.552 × 22.4 = 12.36 L at STP.
Thus, the volume of oxygen gas, O₂ produced is 12.36 L.
